Degree of 3, positive leading coefficient, 2 zeros, 2 turning points. I sketched the condition and the zeros are -1 and 1 and the y-int i made it as 2. How do i make a polynomial function?

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Answer 1 · 2015-09-12T22:45:09

#f(x) = 2x^3-2x^2-2x+2#

#f(x)# will only potentially change sign at the zeros at #x = -1# and #x = 1#, so:

Since the leading coefficient is positive and #f(0) = 2 > 0#, we have:

#f(x) < 0# for #x in (-oo, -1)#

#f(x) > 0# for #x in (-1, 1)#

#f(x) > 0# for #x in (1, oo)#

Due to the degree #3# and #2# zeros, one of the roots must be repeated and from the above it must be the one at #x = 1#

So #f(x) = k(x-1)(x-1)(x+1)# for some #k#

#2 = f(0) = k(-1)(-1)(1) = k#, so #k = 2#

Thus:

#f(x) = 2(x-1)(x-1)(x+1)#

#= 2(x-1)(x^2-1)#

#= 2(x^3-x^2-x+1)#

#= 2x^3-2x^2-2x+2#